 So, what we are going to do today is look at one particular technique that is used for non-linear systems and this is probably one of the first techniques that came in non-linear systems and it is amazing it is a frequency domain based technique, but subsequently of course, with the discovery of circle criterion and so on this particular criterion is more easier to understand, but how so this criterion is called Popov criterion and it was discovered by Popov in 1962 and how the techniques that Popov used of course, are one way to view the Popov criterion and in fact, the Popov criterion can be looked at from several viewpoints. So, what I would like to do is look at the Popov criterion in terms of loop transformations. So, for that we first recall what we did with loop transformations. So, if you recall so, we were always interested in a linear plant and a nonlinearity and they are interconnected in this feedback kind of situation and then of course, without giving any inputs we are interested in knowing whether this feedback circuit is asymptotically stable. Now, of course, this nonlinearity could belong to various kinds of classes and so, we had earlier already said like for example, the nonlinearity could belong to the k infinity class. Now, what we mean by that is that if you draw the nonlinearity characteristics. So, think of the input to the nonlinearity as sigma, output to the nonlinearity as psi, then if you plot sigma against psi, then here I have this line with slope k and then if the nonlinearity is in this sector, what it means is that the nonlinearity is something like that. So, essentially the nonlinearity is something that lies in the first quadrant in between this line and the zeta axis and similarly out here. So, what I am shading that is the area where the nonlinear characteristics lie. Now, going back to the loop if you think of this sigma as the input to the nonlinearity and psi as the output to the nonlinearity, then by the way this is drawn, the input to the linear plant is minus psi and the output of the linear plant is sigma. In other words, G s can be thought of as sigma divided by minus psi. Now, in the loop transformations what is done is this nonlinearity is changed. Now, for a nonlinearity in this class for example, we can convert this into a new nonlinearity in the 0 infinity class and the way we do it is the following. So, what we do is in this particular case we keep the output as it is. So, here we have the nonlinearity and let us say the output is psi, but we modify the input. So, we continue to have the input psi, but what we do is we use a feedback and what for example, we could think of a new input to a nonlinearity sigma 1 and the sigma 1 is given by let us say k times psi minus sigma. So, we think of the new input sigma 1 as being psi minus k sigma. Now, I am sorry. So, in this particular case because it is in the k infinity sector, it is not the psi that we change. So, let me use a new sheet. So, we have the linear plant and the nonlinearity and we have sigma and this is psi. So, this is minus psi, this is sigma and we are assuming that this nonlinearity is in the k infinity sector. That is like saying you have this line of slope k and so the nonlinearity lies in the hatched portion. So, what we do is we keep the input the same, but we change the output. So, how do we do that? We take the same nonlinearity, we keeping the input the same sigma therefore, the output here is psi, but we construct a new output psi 1 which is psi minus k sigma. So, if we have to construct the psi 1 which is psi minus k sigma, then we might as well have a loop. So, k times sigma which we add to psi with a minus sign and so what we have here is psi 1. Now, I am drawing this box here and whatever is inside the box, I call it a new nonlinearity. So, let me call this nonlinearity N L 1. Now, if you think of N L 1, this N L 1 has sigma as its input and psi 1 as its output. Now, if you now look at the characteristics of N L 1 which has sigma as the input and psi 1 as the output. Now, if you look at psi 1 which is psi minus k sigma, then you see this particular nonlinearity N L 1 will lie in the 0 infinity sector. Now, the reason being that of course, the original one was lying in the hatched area. So, the psi could at most given a sigma, the psi could at most be k given a sigma, the psi could at most be k sigma and so if you are doing psi minus k sigma, it could come right down to 0 and so you have the 0 infinity sector. So, this nonlinearity that you have, this nonlinearity is in the 0 infinity sector. Now, the output of this and the input. So, the input is sigma and the output is minus psi and so we will try and maintain a similar configuration. So, this psi 1 is the one which is fed back. Now, if psi 1 is the one which is fed back and we are interested in this G s, we would of course, like the input of G s to be minus psi rather than minus psi 1. So, what we do is we modify the input and you see and we would like the output of the linear part, the output of the linear part is sigma and we would continue that sigma into the input of this new nonlinearity, but the linear plant we would modify in the following way. So, you have sigma, but what we do now is we take the sigma and multiply it by k and we would feed that back with a negative sign in here. So, now if you see and then we would give that input to G s. So, now if you see psi 1, so out here the signal is minus psi 1 and to that from that k sigma is subtracted, but minus psi 1 is minus psi plus k sigma. So, that plus k sigma is cancelled and so what finally, you have in the input of the original linear plant is minus psi which is precisely what you had here. So, the linear plant can be as it was and so by modifying the nonlinearity we also modify the linear plant and therefore, what you have effectively is the same as what you had here. And so the way the, so the new linear plant that you get is what is contained inside this box and let me call it G 1 of s and if you want to look at the transfer function of G 1 of s, G 1 of s has as its output sigma and as its input minus psi 1, for psi 1 I substitute and I get sigma upon minus psi plus k sigma and this then above and below if I divide by minus psi 1, I end up with G upon 1 plus k G. So, this G 1, this new linear plant is related to the old linear plant in the following way. So, G 1 is equal to G 1 plus k G. So, this is the kind of thing that we did. So, I mean, so now we can talk about asymptotic stability of this is the same as the asymptotic stability of this loop, but in this loop because this new nonlinearity is in the zero infinity sector then we know that the linear plant must be a positive real plant. In other words, G 1 s, if G 1 s is positive real then this system is asymptotically stable, but G 1 of s being positive real is saying that as far as this thing is concerned it is asymptotically stable if G upon 1 plus k G is positive real. Now, if you look at it very carefully the original system what we have done in this system is in this loop with the in this arm with the nonlinearity we have fed back with a minus I mean it is a negative feedback with k. Of course, considering that the nonlinearity is going this way this is really a feed forward, but up there we do the same thing we take the sigma multiplied by k and feed it back. So, upstairs it is a negative feedback and here it is a negative feed forward and in this way we managed to convert a nonlinearity in the k infinity sector into a nonlinearity in the zero infinity sector. Now, in the same way we do these other loop transformations. So, for example, had the nonlinearity had the nonlinearity been in the 0 k sector. So, suppose the nonlinearity belong to the 0 k sector. So, that is like saying if you think of this as sigma and this as psi that is sigma and psi this is minus psi this is sigma then this nonlinearity has characteristics which lie in this hatch portion. This is the line with slope k. Now, again our endeavor would be to convert this nonlinearity into a nonlinearity in the zero infinity sector. So, how does one do that? So, in the earlier case we saw that we kept the input the same, but changed the output. In this case one approach that we could take is we could keep the output the same and change the input. So, let us do that. So, you could take this nonlinearity and its output is psi and so the new output will continue to be psi. What we would do is we will create this new nonlinearity and let me call this nonlinearity NL 1 now such that the input is changed. So, let me call the input to this nonlinearity sigma 1 and what is sigma 1? Well, let us take sigma 1 to be sigma minus 1 by k of psi. So, now if you take sigma 1 to be sigma minus 1 by k of psi. So, we feed back from here with a gain 1 by k in here and we add this. Then sigma 1 added to 1 by k times psi will end up in us getting just sigma here. So, the original nonlinearity its characteristics are sort of preserved out here, but what we have done is that the output we have fed back. We have fed back of course, it is a positive feedback such that it cancels this. Now, if we do this, then this nonlinearity. So, why is this nonlinearity now in the 0 infinity sector? Well, you can see that if sigma was the original input and psi was the output, but we modify the input to be sigma minus 1 by k psi, then sigma is here and 1 by k psi. So, suppose the point was let us say out here that means, if it was psi was k times sigma, then what we are saying is that sigma 1 the new input is going to be sigma minus 1 by k of k sigma. That means, sigma 1 is going to be 0. So, you are going to get that same output with 0 as the input. I mean close to 0 as the input. So, one can show that this nonlinearity, this new nonlinearity that you created in NL 1 is in the 0 infinity sector. Now, going back to the other portion, the linear part, if you have minus feedback, so what you have here is minus psi. So, we could put the linear plant there. So, if you put the linear plant here from here, you know that with minus psi as the input, sigma must be the output, but we want sigma 1 to be the input here. So, we have to modify this. So, what we do is we take this minus psi and multiply it by 1 by k and add it to the sigma. So, then effectively what you have here is sigma 1. So, this is the feedback that you will have. So, now, this here is the linear part here g 1 of s and so g 1 of s interconnected to this new nonlinearity NL 1 is equivalent to g of s interconnected to this original nonlinearity NL. Now, if you look at the arrangement that we have done, what we have done here is of course, here of course, the signal is flowing this way, but what we have done is we have taken the psi here and multiplied it by again 1 by k and put it back here with a positive sign and so here also we will do the same thing. So, just like in the last case, we do exactly similar things in the upper and the lower. Of course, what it means is here it is a positive feedback and here it is a positive feed forward. Now, in this case of course, this new g 1 of s turns out to be nothing but g of s plus 1 by k. And then therefore, for this system to be asymptotically stable, what we require is that this new plan g 1 of s because this nonlinearity is in the zero infinity sector, we want this g 1 of s to be positive real. In other words, g of s plus 1 by k, this whole thing must be positive real. Of course, this then you could interpret as saying that the Nyquist plot of g of s should lie to the right of minus 1 by k. Now, notice that in the last case and in this case, in both these cases, what you have done is some sort of feedback, feed forward that you have done and whatever you do in this arm which contains the nonlinearity, you do a similar thing up there with the linearity. And what it effectively does is it neutralizes the effect of that thing in some way so that the original nonlinearity and the original linearity still maintain the relation that they were supposed to maintain. Now, the point is that the gains that we used in both these situations were gains that were constant gains, but perhaps we could introduce gains maybe in such a way or maybe in series with the linear plant. For example, in series with the nonlinearity or in series with the nonlinearity and therefore, in series with the linear plant and so on and modify the loops. So, let us just try out one particular example. So, for example, if the original plant is like that with the nonlinearity like this, now let us not bother about where the nonlinearity lies, but let us do the following. So, the suppose the nonlinearity is here and let us assume that the output to the nonlinearity is kept the same. So, the output is kept as psi, but what we do is input. So, one would want the input here to be sigma, but maybe we put a gain here and maybe the gain that one puts here is let us say 1 by k. So, what that would mean is out here the signal that you have is k sigma and we could think of unity feedback. So, just like in the last case, but I have pushed the gain from 1 by k from the feedback loop in here. So, then we could think of the new input as sigma 1 and so in this case you have sigma 1 is k sigma minus psi and so this here is our nonlinearity. Then in order to find the equivalent thing to equivalent interconnection to this interconnection you have a gs here. So, this psi, so this minus psi, so the output of this is sigma, but you want to get sigma 1, sigma 1 is obtained this way. So, what we can do is multiply this by k. So, therefore you get k sigma and then you take this minus psi and feed it forward and therefore, what you will get is sigma 1 there. Now, the construction that I have done here is exactly the same as the earlier construction that I did. So, this construction and this construction are exactly the same. The only thing is that this gain I pushed it in here and so therefore, this gain has to get pushed in here, but when I do that here it is 1 by k there when it gets pushed it gets pushed to k and that is essentially because here it is in the feedback arm and there it is in the feed forward arm. So, when you put it in series it becomes k whereas, from the feedback arm it remains as it is and so what we have effectively done is we have done I mean we could just assume that this nonlinearity here is really a nonlinearity in the 0 k sector and therefore, this new nonlinearity NL 1. So, NL 1 would then be in the 0 infinity sector and this is the original plan G s and the new plan G 1 of s. The transfer function for G 1 of s is 1 plus k times G s which of course is roughly the same as the result that we had got earlier only it is this new G 1 of s is going to be k times that old G 1 of s, but you know by just multiplying by a scalar k you do not change anything. Now, instead of having gains here maybe we could have transfer functions. So, one kind of transfer function that you could possibly have is if you think you have G s here and you have a nonlinearity and let us again assume that this nonlinearity is in the 0 k sector and what we do is a construction similar to the earlier construction. So, what we do is this nonlinearity the k. So, this sigma this is psi therefore, you have the minus psi here sigma there and the nonlinearity is in the 0 k sector which essentially means this hatched area is where you have the nonlinearity this is sigma this is psi. So, what we do is this nonlinearity we multiply by first order transfer function like this. So, if you multiply the nonlinearity by a first order transfer function like this and of course this nonlinearity is the same as this nonlinearity. So, the signals across you want to maintain to be the signals across there. So, you have sigma here and psi here and so, if you use this psi in the feedback. So, you have minus psi here. So, let us assume you have a transfer function G s here. So, what you would get here is sigma and let me do the following. Let me assume that there is a gain here which is 1 by k of the psi and I feed it back that way. So, if you have something like this you should do a similar thing there. So, and I have this new input sigma 1 and so, this is my modified nonlinearity nL1. So, now, one would like to do something out there such that this and that sort of matches up. So, what we do up there is I can again here I have a feedback positive feedback loop. So, there is a plus sign here and a gain of 1 by k. So, I take this with a gain of 1 by k and I am going to add it here with a plus sign and earlier we saw that if you had if you had 1 by k here then you have k here. So, we will just blindly do that and we will see what happens. So, let us blindly do that. So, the reciprocal of this is 1 plus AS. So, let us put that gain 1 plus AS out there and let me call this output that comes here sigma 1 and connect it up here. So, maybe this is looking all to clamped up. So, maybe I will use a new sheet with this particular diagram. So, we had the nonlinearity and then we used 1 upon 1 plus AS here and again a 1 plus k which we fed back in here with a positive sign and this whole thing is the new nonlinear system NL1 let me call it and so NL1's output is psi input is sigma 1 this is sigma here. We are assuming this is sigma here we do not yet know, but let us just do the construction and see whether we get something meaningful G s. So, with the input minus psi. So, the output here must be sigma and then we have gain like that and what we do is a similar construction. So, 1 by k through here again with a positive sign. Now, from the earlier situation here we know G of s is sigma by minus psi. So, let us just make this assumption. So, assuming that the output of the nonlinearity is still psi which is fed back here. So, you have minus psi here. So, this must be sigma. Now, if sigma passes through here what you get here is going to be a times sigma dot plus sigma and then when you have this thing. So, what you are going to get is minus 1 by k psi and this whole thing is sigma 1. Let us assume this whole thing is sigma 1. Now, if this whole thing is sigma 1. So, sigma 1 which is coming in here you have psi 1 by k times psi added to it. So, what you have here the signal here is going to be a times sigma dot plus sigma and if you have the signal a times sigma dot plus sigma here 1 upon 1 plus AS acting on that will in fact give you the signal sigma. So, this whole thing looks consistent. So, this whole linear plant that you have let us call that G 1 of s and this whole new nonlinearity N L 1. Then perhaps we could say that this interconnection is asymptotically stable and if we can say that this interconnection is asymptotically stable that would be equivalent to going back to the original plant and saying this interconnection is asymptotically stable. Now, in here if you look at this linear plant G 1 s this linear plant G 1 s is given by 1 plus AS times GS plus 1 by k. Now, if this linear plant is positive real. So, now let us make some assumptions. So, if this is passive and this is passive then the interconnection of two passive systems is passive. So, I mean this being passive and this being passive the interconnection is passive. So, because each one of them is passive one could find a Lyapunov function or a storage function for each one of them and the storage functions. So, recall that for passivity if you have a passive system then you have a storage function V such that V dot is less than input times output. This is how we sort of less than equal to input times output this is how we characterize passivity. So, what we can start doing is the following. So, let us assume that this whole plant this whole new linear plant G 1 of s is passive. Now, this whole thing being passive is like saying that G 1 of s which is 1 plus AS times GS plus 1 by k this is positive real. And then we have gone through several of these theorems earlier like the Kalman-Yakobovic lemma by which one can find for this part a storage function V 1 such that V 1 dot is less than equal to the input times the output which in this case is minus sigma times psi. So, if you can also show that this guy is passive then the sum of these two is going to be passive. So, how to show that this particular transfer function is also passive. Now, for that what we will do is we will look at the differential equation. So, the differential equation that one could look at is the following this whole nonlinearity is there. So, let me write it out in the following way. So, let me call the nonlinearity f such that if sigma is the input then psi is the output and what I am really saying is psi is equal to f of sigma. And then earlier we had 1 upon 1 plus AS here and then we also had this feedback gain of 1 by k. Now, if this is sigma then of course from the last slide we saw that this must be this signal here must be sigma plus a times sigma dot. And because of this we know that sigma 1 must be given as sigma plus a times sigma dot plus 1 by k times psi. Now, let me write this psi using the nonlinearity thing let me write this psi as f of sigma and so I have sigma 1 is equal to sigma plus a times sigma dot plus 1 by k f of sigma. This thing is really like a differential equation with so think of this as a differential equation with sigma as the state variable and sigma 1 as the input. So, one could write something like a state space equation saying a sigma dot is equal to minus sigma minus 1 by k f of sigma plus sigma 1 where sigma 1 is thought of as the input and sigma is the state and then of course one could think of the output equation as psi is equal to f of sigma. So, take this equation this differential equation where sigma is a state sigma 1 is the input and this has the output equation. So, these two equations you have. Now, let us make one more assumption let us make the assumption that this nonlinearity this nonlinearity f is in the 0 k sector k is related to this k where the gain was 1 by k. So, the f the characteristics of f lie in the hatched area. Now, if this is the case then I sort of decide that I will take Lyapunov function in the following way you see if the characteristic of this f. So, you see if the characteristic of the f of the function f so we have this differential equation a sigma dot is equal to minus sigma sorry minus 1 by k f of sigma plus sigma 1 and you have psi is equal to f of sigma that is the system and this f is such. So, if this is the slope k sigma psi f is such that f looks something like that. Now, so this is the curve f of sigma then of course it would be clear that if you take integral of f of sigma d sigma going from 0 to some point x then this quantity is always going to be positive because if that x is somewhere here then this integral is positive. If it is somewhere here then this integral this integral is negative but of course the limits are going the other way. So, when you change the limit it becomes positive. So, you get something like this. So, this is going to be greater than 0 for all x. So, I have been using sigma as a state variable. So, maybe I could just rewrite this as x a x dot equal to minus x 1 by k f of x plus u let me call it u the input and then I am saying y is equal to f of x this is the state space equations is the same equations as before and we see that this function is such that if you take the integral from 0 to x this is true. So, I propose that we use a Lyapunov function v x given as this is equal to a times integral from 0 to x of f sigma d sigma. Therefore, this Lyapunov function or this Lyapunov function candidate has the property that this is greater than 0 for all x greater than 0. So, now if you evaluate v dot of x. So, v dot of x from this evaluation is going to be a times f of x times x dot. Now, a times x dot we can substitute in this. So, you get minus x times f of x so first term and then you get minus 1 by k f of x squared that is that term and then you get plus f x u. Now, if you take any x x times f x f x is. So, for positive x f x is positive for negative x f x is negative. So, this quantity here x times f x is positive. So, minus of x times f x is negative. Similarly, this quantity here is also negative and therefore, one can conclude that v dot x is less than f x times u, but f x is y. So, u times y. So, this system that you have the state space system that you have with this particular Lyapunov function you can see or storage function you can see that this is passive. So, if you go back here this system that we have drawn here the system is passive with the storage function given as v is equal to a integral 0 to sigma f of r dr. Let me call this v 1 with this storage function and we already saw that in this interconnection that we were looking at this portion because, it is positive real has a storage function v 1 has a storage function v 1. So, is that v 1 dot is less than equal to minus sigma psi because the output here is sigma and the input is minus psi. And out here if you take this let me call it v 2 then we know v 2 dot is less than is less than equal to the input sigma 1 times the output psi sorry I have to correct one thing here. Here the output is sigma 1 and not sigma and so is minus sigma 1 psi. So, now if you take the Lyapunov function if you take the Lyapunov function of that net system to be v 1 plus v 2 which is equal to okay. So, that v 1 would have been given by because that that linear plant that effective linear plant that you had. So, this is the v 1 this effective linear plant you could look at the state space realization of this. And from that you could write in terms of the states of this thing as some x transpose p x where this p will satisfy all the Kalman Yakovovich Lemma conditions and plus a times integral 0 to x of f of r dr this as the net as the net Lyapunov function v then you have v dot equal to v 1 dot plus v 2 dot which of course we can calculate. But we know that v 1 dot this is less than v 1 dot we know is less than minus sigma 1 psi v 2 dot is less than sigma 1 psi. So, some of these two is 0. So, v this thing that you had this Lyapunov function is such that v dot is less than 0. And this quantity here is going to be positive because this quantity is going to be positive and this p has been obtained to be a positive definite matrix. Therefore, by Lyapunov theorem we have asymptotic stability. So, what we are effectively saying so this is what is called the Popov criterion and what the Popov criterion is saying is the following that if you had a linear plant and you had a nonlinearity and you had it interconnected in this way and this nonlinearity was in the 0 k sector and the linear plant was such that 1 plus A s times G s plus 1 by k is positive real then this interconnection is going to be asymptotically stable. Of course, this nonlinearity must be such that it is a memory less nonlinearity. The memory less Lipschitz nonlinearity, but this is the effective result of the Popov criterion. So, it looks like I am out of time now. So, let me stop for now.